MATH383: A first course in differential equationsJanuary 16, 2019

Due date: Jan 23, 2019, 11:55PM.

Bibliography: Trench Chap. 2

1. Exercises 6-9, p. 41

   Ex 6

   $y^{\text{'}}+(1+x)y/x=0$, $y\left(1\right)=1.$ Assume $x\ne 0$, and solve by integration

   ${\displaystyle \frac{y^{\text{'}}}{y}=-\frac{x+1}{x}\Rightarrow \int \frac{dy}{y}=-\int \frac{x+1}{x}dx\Rightarrow \ln y-\ln c=-(x+\ln x)\Rightarrow y=\frac{c}{x}{e}^{-x}.y\left(1\right)=\frac{c}{e}=1\Rightarrow c=e}$

   Solution is $y=e^{1-x}/x.$ Check in Maxima

   (%i4)

   ode2('diff(y,x)/y=-(x+1)/x,y,x);

   ${\displaystyle \text{(\%o4)}y=\frac{\%c{\mathrm{e}}^{-x}}{x}}$

   (%i5)

   ic1(%,x=1,y=1);

   ${\displaystyle \text{(\%o5)}y=\frac{{\mathrm{e}}^{1-x}}{x}}$

   Ex 7

   $x{y}^{\text{'}}+(1+1/\ln x)y=0,y\left(e\right)=1$. Assume $x\ne 0$, and solve by integration

   ${\displaystyle \frac{y^{\text{'}}}{y}=-\frac{1}{x\ln x}-\frac{1}{x}\Rightarrow \ln y-\ln c=-\int (\frac{1}{x\ln x}+\frac{1}{x})dx=-\ln (\ln x)-\ln x=-\ln (x\ln x)\Rightarrow y=\frac{c}{x\ln x}}$ ${\displaystyle y\left(e\right)=\frac{c}{e}=1\Rightarrow y=\frac{e}{x\ln x}.}$

   Verify in Maxima:

   (%i9)

   ode2('diff(y,x)*x+(1+1/log(x))*y=0,y,x);

   ${\displaystyle \text{(\%o9)}y=\frac{\%c}{x\log \left(x\right)}}$

   (%i10)

   ic1(%,x=%e,y=1);

   ${\displaystyle \text{(\%o10)}y=\frac{\mathrm{e}}{x\log \left(x\right)}}$

   (%i11)

   Ex 8

   $x{y}^{\text{'}}+(1+x\cot x)y=0,y(\pi /2)=2$. Assume $x\ne 0$, and solve by integration

   ${\displaystyle \frac{y^{\text{'}}}{y}=-\frac{1}{x}-\frac{1}{\cot x}\Rightarrow \ln y-\ln c=-\ln x-\ln \sin x\Rightarrow y=\frac{c}{x\sin x}}$ ${\displaystyle y(\pi /2)=2c/\pi =2\Rightarrow c=\pi {.}^{}y=\frac{\pi }{x\sin x}}$

   Verify in Maxima

   (%i16)

   ode2('diff(y,x)*x+(1+x*cot(x))*y=0,y,x);

   ${\displaystyle \text{(\%o16)}y=\frac{\%c}{x\sin \left(x\right)}}$

   (%i17)

   ic1(%,x=%pi/2,y=2);

   ${\displaystyle \text{(\%o17)}y=\frac{\pi }{x\sin \left(x\right)}}$

   (%i18)

   Ex 9

   $y^{\text{'}}+ky/x=0,y\left(1\right)=3$. Integrate:

   ${\displaystyle \frac{y^{\text{'}}}{y}=-\frac{k}{x}\Rightarrow \ln y-\ln c=-k\ln x\Rightarrow y=c{x}^{-k},y\left(1\right)=c=3.y\left(x\right)=3x^{-k}}$

   (%i22)

   ode2('diff(y,x)+k*y/x=0,y,x);

   ${\displaystyle \text{(\%o22)}y=\%c{\mathrm{e}}^{-k\log \left(x\right)}}$

   (%i23)

   ic1(%,x=1,y=3);

   ${\displaystyle \text{(\%o23)}y=3{\mathrm{e}}^{-k\log \left(x\right)}}$

   (%i24)

2. Exercises 12-15, p.41

   Ex 12

   $y^{\text{'}}+3y=1$. Solve by integration

   ${\displaystyle y^{\text{'}}=1-3y=\frac{dy}{dx}\Rightarrow dx=\frac{dy}{1-3y}\Rightarrow x+c=-\frac{1}{3}\ln (1-3y)}$

   (%i1)

   ode2('diff(y,x)=1-3*y,y,x);

   ${\displaystyle \text{(\%o1)}y={\mathrm{e}}^{-3x}(\frac{{\mathrm{e}}^{3x}}{3}+\%c)}$

   (%i3)

   plotdf(1-3*y);

   (%i4)

   

   Ex 13

   $y^{\text{'}}+(1/x-1)y=-2/x$. Use variation of parameters. Homogeneous equation $y^{\text{'}}+(1/x-1)y=0$ has solution:

   ${\displaystyle \frac{y^{\text{'}}}{y}=1-\frac{1}{x}\Rightarrow \ln y-\ln c=x-\ln x\Rightarrow y_h=e^x/x}$

   Seek solution of form $y=u{y}_h\Rightarrow u^{\text{'}}=-2e^{-x}\Rightarrow u=2e^{-x}+c\Rightarrow y=(2+c{e}^x)/x$. Verify in Maxima

   (%i3)

   ode2('diff(y,x)+(1/x-1)*y=-2/x,y,x);

   ${\displaystyle \text{(\%o3)}y=\frac{(2{\mathrm{e}}^{-x}+\%c){\mathrm{e}}^x}{x}}$

   (%i5)

   plotdf(-2/x-(1/x-1)*y,[x,.1,4.1],[y,-2,2]);

   

   (%i6)

   Ex 14

   $y^{\text{'}}+2xy=x{e}^{-x^2}$. Solve by variation of parameters. Homogeneous equation solution

   ${\displaystyle \frac{y^{\text{'}}}{y}=-2x\Rightarrow y_h=e^{-x^2}(\mathrm{particular}\mathrm{solution})}$

   Variation of parameters $y=u{y}_h\Rightarrow u^{\text{'}}=x\Rightarrow u=x^2/2+c\Rightarrow y=(x^2/2+c)e^{-x^2}$. Verify in Maxima

   (%i7)

   ode2('diff(y,x)+2*x*y=x*exp(-x^2),y,x);

   ${\displaystyle \text{(\%o7)}y=(\frac{x^2}{2}+\%c){\mathrm{e}}^{-x^2}}$

   (%i9)

   plotdf(-2*x*y+x*exp(-x^2),[x,-2,2],[y,-2,2]);

   

   Ex 15

   $y^{\text{'}}+2xy/(1+x^2)=e^{-x}/(1+x^2)$. Solve by variation of parameters. Homogeneous solution:

   ${\displaystyle \frac{y^{\text{'}}}{y}=-\frac{2x}{1+x^2}\Rightarrow y_h=\frac{1}{1+x^2}.}$

   Seek solution $y=u{y}_h\Rightarrow$

   ${\displaystyle u^{\text{'}}=e^{-x}\Rightarrow u=-e^{-x}+c\Rightarrow y=\frac{c-e^{-x}}{1+x^2}}$

   Verify in Maxima

   (%i14)

   ode2('diff(y,x)+2*x*y/(1+x^2)=exp(-x)/(1+x^2),y,x);

   ${\displaystyle \text{(\%o14)}y=\frac{\%c-{\mathrm{e}}^{-x}}{x^2+1}}$

   (%i15)

   plotdf(-2*x*y/(1+x^2)-exp(-x)/(1+x^2),[x,-2,2],[y,-2,2]);

   ${\displaystyle }$

   (%i16)

3. Exercises 30-33, p.42

   Ex 30

   ${\displaystyle (x-1)y^{\text{'}}+3y=\frac{1}{{(x-1)}^3}+\frac{\sin x}{{(x-1)}^2},y\left(0\right)=1.}$

   Solve by variation of parameters. Homogeneous solution

   ${\displaystyle \frac{y^{\text{'}}}{y}=-\frac{3}{x-1}\Rightarrow \ln y=-3\ln (x-1)\Rightarrow y_h=\frac{1}{{(x-1)}^3}.}$

   Seek solution as $y=u{y}_h$.

   ${\displaystyle u^{\text{'}}=\frac{1}{x-1}+\sin x\Rightarrow u=\ln (x-1)-\cos x+c\Rightarrow }$ ${\displaystyle y=\frac{\ln (x-1)-\cos x+c}{{(x-1)}^3},y\left(0\right)=1-c=1\Rightarrow c=0}$

   Verify

   (%i6)

   y30: (log(x-1)-cos(x))/(x-1)^3

   ${\displaystyle \text{(\%o6)}\frac{\log (x-1)-\cos \left(x\right)}{{(x-1)}^3}}$

   (%i11)

   fullratsimp((x-1)*diff(y30,x)+3*y30);

   ${\displaystyle \text{(\%o11)}\frac{(x-1)\sin \left(x\right)+1}{x^3-3x^2+3x-1}}$

   (%i12)

   partfrac(%,x);

   ${\displaystyle \text{(\%o12)}\frac{\sin \left(x\right)}{{(x-1)}^2}+\frac{1}{{(x-1)}^3}}$

   (%i13)

   Ex 31

   $x{y}^{\text{'}}+2y=8x^2,y\left(1\right)=3$. Solve by variation of parameters. Homogeneous solution:

   ${\displaystyle \frac{y^{\text{'}}}{y}=-\frac{2}{x}\Rightarrow y_h=x^{-2}}$

   Try $y=u{y}_h\Rightarrow u^{\text{'}}=8x^3\Rightarrow u=2x^4+c\Rightarrow y=(2x^2+c/x^2).$ Initial condition $y\left(1\right)=2+c=3\Rightarrow c=1$, and solution is $y=2x^2+c/x^2$. Verify in Maxima

   (%i13)

   ode2('diff(y,x)*x+2*y=8*x^2,y,x);

   ${\displaystyle \text{(\%o13)}y=\frac{2x^4+\%c}{x^2}}$

   (%i14)

   ic1(%,x=1,y=3);

   ${\displaystyle \text{(\%o14)}y=\frac{2x^4+1}{x^2}}$

   (%i15)

   partfrac(%,x);

   ${\displaystyle \text{(\%o15)}y=2x^2+\frac{1}{x^2}}$

   (%i16)

   Ex 32

   $x{y}^{\text{'}}-2y=-x^2,y\left(1\right)=1$. Solve by variation of parameters. Homogeneous solution

   ${\displaystyle \frac{y^{\text{'}}}{y}=\frac{2}{x}\Rightarrow y_h=x^2.}$

   Try $y=u{y}_h\Rightarrow u^{\text{'}}=-1/x\Rightarrow u=-\ln x+c\Rightarrow y=(c-\ln x)x^2$. Initial condition: $y\left(1\right)=c=1$. Solution:

   ${\displaystyle y=(1-\ln x)x^2.}$

   Verify:

   (%i16)

   ode2('diff(y,x)*x-2*y=-x^2,y,x);

   ${\displaystyle \text{(\%o16)}y=x^2(\%c-\log \left(x\right))}$

   (%i17)

   ic1(%,x=1,y=1);

   ${\displaystyle \text{(\%o17)}y=x^2-x^2\log \left(x\right)}$

   (%i18)

   Ex 33

   $y^{\text{'}}+2xy=x,y\left(0\right)=3$. By variation of parameters. Homogeneous solution:

   ${\displaystyle \frac{y^{\text{'}}}{y}=-2x\Rightarrow y_h=e^{-x^2}.}$

   Try $y=u{y}_h\Rightarrow u^{\text{'}}=x{e}^{x^2}\Rightarrow u=\frac{1}{2}{e}^{x^2}+c\Rightarrow y=(\frac{1}{2}+c{e}^{-x^2}),y\left(0\right)=\frac{1}{2}+c=3\Rightarrow c=\frac{5}{2}\Rightarrow$

   ${\displaystyle y=\frac{1}{2}(1+5e^{-x^2}).}$

   Verify:

   (%i18)

   ode2('diff(y,x)+2*x*y=x,y,x);

   ${\displaystyle \text{(\%o18)}y={\mathrm{e}}^{-x^2}(\frac{{\mathrm{e}}^{x^2}}{2}+\%c)}$

   (%i19)

   ic1(%,x=0,y=3);

   ${\displaystyle \text{(\%o19)}y=\frac{{\mathrm{e}}^{-x^2}({\mathrm{e}}^{x^2}+5)}{2}}$

   (%i20)

4. Exercises 1-4, p. 52

   Ex 1

   $y^{\text{'}}=(3x^2+2x+1)/(y-2)$. Assume $y\ne 2$, and integrate

   ${\displaystyle \int (y-2)dy=\int (3x^2+2x+1)dx\Rightarrow \frac{1}{2}{(y-2)}^2=x^3+x^2+x+c.}$

   There are two solutions

   ${\displaystyle y=2\pm \sqrt{2(x^3+x^2+x+c)}}$

   Ex 2

   $(\sin x)(\sin y)+(\cos y)y^{\text{'}}=0$. Assume $\sin y\ne 0$, divide by $\sin y$ and integrate

   ${\displaystyle \int \cot ydy=-\int \sin xdx\Rightarrow \ln (\sin y)=\cos x+c\Rightarrow \sin y=e^{c+\cos x}.}$

   Solutions will only exist in intervals such that $0<e^{c+\cos x}⩽1\Rightarrow c+\cos x⩽0$. In such intervals there are infinitely many solutions $y={(-1)}^k\arcsin e^{c+\cos x}+k\pi ,k\in \mathbb{Z}$. Rewriting the DE as $y^{\text{'}}=f(x,y)=-\sin x\tan y$, notice that $f$ is not continuous at $\pm k\pi /2$.

   Ex 3

   $x{y}^{\text{'}}+y^2+y=0$. For $x\ne 0$, rewrite to obtain a Bernoulli equation $y^{\text{'}}+y/x=-y^2/x$. Solve homogeneous equation

   ${\displaystyle \frac{y^{\text{'}}}{y}=-\frac{1}{x}\Rightarrow y_h=\frac{1}{x}.}$

   Try $y=u{y}_h\Rightarrow u^{\text{'}}{y}_h=-{(u{y}_h)}^2/x\Rightarrow$

   ${\displaystyle \frac{u^{\text{'}}}{u^2}=-\frac{y_h}{x}=-\frac{1}{x^2}\Rightarrow \frac{1}{u}=c-\frac{1}{x}\Rightarrow u=\frac{x}{cx-1}\Rightarrow y=\frac{1}{cx-1}}$

   (%i23)

   y3: 1/(c*x-1);

   ${\displaystyle \text{(\%o23)}\frac{1}{cx-1}}$

   (%i24)

   x*diff(y3,x)+y3^2+y3

   ${\displaystyle \text{(\%o24)}\frac{1}{cx-1}-\frac{cx}{{(cx-1)}^2}+\frac{1}{{(cx-1)}^2}}$

   (%i25)

   fullratsimp(%)

   ${\displaystyle \text{(\%o25)}0}$

   (%i26)

   Ex 4

   $y^{\text{'}}\ln \left|y\right|+x^{2}y=0$. For $y>0$ obtain

   ${\displaystyle \frac{y^{\text{'}}\ln y}{y}=-x^2\Rightarrow \frac{1}{2}{(\ln y)}^2=-\frac{1}{3}{x}^3+\frac{1}{2}c}$

   There are two solutions

   ${\displaystyle y=\exp [\pm \sqrt{c-\frac{2}{3}{x}^3}].}$